Solution Manual for Complete Business Statistics 7th Edition Aczel

download full file at http://testbankcafe.com CHAPTER 2 PROBABILITY 2-1. Objective and subjective. 2-2. An event is a set of basic outcomes of an experiment. The union of two events is the set containing all basic outcomes that are either in one event or in the other, or in both. The intersection of two events is the set of basic outcomes that are members of both events. 2-3. The sample space is the universal set pertinent to a given experiment. It is the set of all possible outcomes of an experiment. 2-4. The probability of an event is a measure of the likelihood of the occurrence of the event. When sample points are equally likely, the probability of the event is the relative size of the set comprising the event within the sample space. 2-5. The union G  F is the event that the baby is either a girl, or is over 5 pounds (of either sex). The intersection G  F is the event that the baby is a girl over 5 pounds. 2-6. The union is the event that the player scores in the game with A, or in the game with B, or in both. The intersection is the event that the player scores in both games. 2-7. Sample Space first toss second toss > first There are 36 possible outcomes tossing two dice. There are 15 possible outcomes where the second toss is greater than the first. P(Second Toss > First) = 15/36 = 0.417 First Toss Second Toss > First 1 2 3 4 5 6 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6 none download full file at http://testbankcafe.com 11.com .10 = 0.06 .035 c. the probability of winning is 1/37 = 0.2778 2. S  B : purchase stock or bonds. 2-18. 2. Let 0. 2-20. P (T  R )  P (T )  P ( R )  P (T  R ) = 0. down (respectively).1667 P(nonfunctioning data device and nonfunctioning cell phone) = 1/18 = 0.99 2-19.015  0.85. b. The team is very likely to win. 2-17. P(nonfunctioning data device) = 5/18 = 0.0 = 0. a. 2-10.49 download full file at http://testbankcafe. R  T is the event that a randomly chosen person is exposed to the ad on the radio or the ad on television. P(Public Sector) + P(Corporate Sector) – P(both sectors) = 0. better odds (for the player) than the American version. Since there are only 37 possible outcomes. More likely to occur than not to occur.02 = 0. P(F) + P(>50) – P(F & >50) = 12/20 + 2/20 – 2/20 = 0. The two events are mutually exclusive.015 . or both. P(first shopper detected) + P(second detected) .12 2-12.985 . P(D )  I  P(D)  I  0. R  T is the event that a randomly chosen person is exposed to the ad on the radio and the ad on television.0. 0. 2-15.34 – 0. S  B : purchase stock and bonds.6 P(< 30) = 2/20 = 0. for example. or both.98 + 0. Because of this.0556 P(nonfunctioning data device or nonfunctioning cell phone) = 0.93 = 0.0556 = 0. D be the events: machine is out-of-control.25 + 0. Then we need P(O  D)  P(O)  P(D)  P(O  D) = 0. This event and D are mutually exclusive.download full file at http://testbankcafe.08 + 0. 2-16.94 .13.027.1667 – 0.02 + 0.3889 2-14.000 population.1 2-21. Based on murder statistics for a given time period: total murders per 100. is a typical "very likely" probability.0. 2-9.com 2-8.P(nonfunctioning cell phone) = 3/18 = 0.2778 + 0. the house admission fee makes sense.P(both detected) = 0. P(D | L) = 0.32 Given that P(M  R) = 0.34 2-28.2778 P(functioning data device) = 1 – P(nonfunctioning data device) = 1 – 0.2778 = 0.5% of the packages are late. c. Given that P(A | H) = 0.454 i.2065 Mutually exclusive events: P(M  S) = P(M) + P(S) = (198 + 968) / 1976 = 0.25 and P(D) = 0. P(VT  CE) = 380/550 + 412/550 .90 2-23.40.94 and P(H) = 0. and 4%.357/550 = 0.12 / .611 download full file at http://testbankcafe.10) = 0. Given that P(R | B) = 0.60 2-30.025 2-31.28 = 0.85 and P(B) = 0.451 f.34 P(M | R) = P(M / R) P(R) = (0. P(R  B) = P(R | B) P(B) = (. P(E  O) = P(E) + P(O) – P(E  O) = (408 + 590 – 100) / 1976 = 0.download full file at http://testbankcafe. P(M) = 198 / 1976 = 0.59 P(H) = 284/1976 = 0.10.144 P(S | H) = P(S  H) / P(H) = (128 / 1976) / (284 / 1976) = 0. P(D | L) = P(L  D) / P(L) = . From (2-12): P(nonfunctioning data device) = 0. the answer is 2(2+2+4) = 25% 2-27.1002 P(E) = 408/1976 = 0. P(P | E) = P(P  E) / P(E) = (233 / 1976) / (408 / 1976) = 0.94)(.7909 2.40) = 0.571 g. b.com .7222 2-25.85 + 0.1% successfully completed. P(W | O) = P(W  O) / P(O) = (99 / 1976) / (590 / 1976) = 0.12 and P(L) = 0.65.80)(0.80 and P(R) = 0.20.33 . 25%: Reading the market shares as 2 %. P(A  H) = P(A | H) P(H) = (. e. P(M  H)  P(M)  P(H)  P(M  H) = (11 + 8 – 5) / 28 = 1 / 2 = 0.20 = 0.60 Given that P(L  D) = 0. 2. d.24.569 2-32.com 2-22. P( S  B )  P( S )  P( B)  P( S  B ) = 0.65) = 0. 2%.4. Given that P(N | D) = 0.0.40) = 0.25)(. 61. P(M | R) = 0.168 h.50 2-26. P(H  S) = P(H) + P(S) – P ( H  S ) = (284 + 968 – 128) / 1976 = 0.85)(. P(N  D) = P(N | D) P(D) = (.32 2-29. 0. a. (Use template: Probability of at least 1.634 g.35)5 = 0.com .99475.65 0.download full file at http://testbankcafe.65 0.455 c.455 + 0. P(I | D) = P(I  D)/P(D) = (85 / 246) / (134 / 246) = 0.65 Prob. 0. P(D) = 112/ 246 = 0.S denote the events: top Executive made over $1M. Let E. of at least one success 0.65 0. P( S ) = 3 / 10 = 0.484 = 0.00525 = 0. a. Probability of at least one success from many independent trials. Then: a.801 2-34. The probability that none of the five people chosen is obese would be (0. P(D  I) = P(D) + P(I) – P ( D  I ) = 0. P(I  D) = 34 / 246 = 0. the probability that the person is obese is 65% and the probability that the person is not obese is 35%.333 2. P(E | S ) = P(I  S )/P( S ) = (2 / 10) / (3 / 10) = 2/3 = 0. Shareholders made money. Success Probs 1 2 3 4 5 2-37. P(E) = 3 / 10 = 0.65 0.com 2-33.xls) For any single person.484 b. respectively.138 / 0.285 f.35. P(D | I) = P(D  I) / P(I) = 0. P(S | E) = P(S  E)/P(E) = (1/10)/ (3/10) = 1/3 = 0.138 = 0.30 b.90 = 0. The probability that at least one of the people chosen is overweight and obese is: P(≥ 1 overweight & obese) = 1 – P(none are overweight and obese) = 1 – 0.00525. P(I  D) = 49/ 246 = 0. P(Critical Danger | Dependence on Forested Areas) = P(CD ∩ DFA) / P(DFA) = 0.484 – 0.333 2-36. P(I) = 119 / 246 = 0.199 e.667 d.138 d.30 c.30 / 0.9947   2   2   2   5   5   19   19   P(at least one job) = 1 – P(no jobs) = 1 –                (by independence) =   3   3   3   6   6   20   20   0.8143 download full file at http://testbankcafe. P(increase sales all three countries) = P(increase sales in US) x P(increase sales Australia) x P(increase sales in Japan) P(increase sales all three countries) = (0.90)(0.93)(1 – 0.download full file at http://testbankcafe.75 0.com 0.6 Prob.05 0. of at least one success download full file at http://testbankcafe.8142 2-38.333 0.9900 .00945 = 0.99055 2-42.998 2-41.333 0.167 0.167 0.88)(1 – .70) = 1 .02)(0.9 0.333 0. P(device works satisfactorily) = 1 – P(both components fail) = 1 – (0.0. Success Probs 1 2 3 4 5 6 7 0.xls) Probability of at least one success from many independent trials.91) = 0.com (Use template: Probability of at least 1.85) = 0. (Use template: Probability of at least 1.1) = 1 – 0.72675 2-40. Assume independence: P(at least one arrives on time) = 1 – P(all three fail to arrive) = 1 – (1 – . P(at least one benefit) = 1 – P(no benefits) = 1 – (1 – 0.95)(0.55)(1 – 0.05 Prob. of at least one success 0.xls) Probability of at least one success from many independent trials. Success Probs 1 2 3 0.90)(1 – .002 = 0.99892 2-39. )) = 1 – (1 – (110/2100))(1 – (200/600))(1 – (. 2-46.82 0.2202 P(D  I) = 34/246 = 0.999904.com 2-43. Therefore. The probability of not dying in a car crash for the next five years. of at least one success 0. the probability of dying in a car crash over the next five years is: 1 – P(not dying) = 1 – 0. The device works if at least one out of three works.9995) = 0.99928 2-49. Success Probs 1 2 3 4 2-44. P(D)P(I) = (112/246)(119/246) = 0.99952.20  0.91)(1 – 0.80) = 0.96)(1 – 0.01440 The two events are not independent (but they are close) 2-45.s.xls) The probability of dying in a car crash in France in 2003 is 5732 / 59625919 = 0. The probability on not dying in a car crash is 0.01440 P(H  M) = 29/1976 = 0.09 The two events are not independent.7 0.)P(not getting schist.99952 = 0.9989 P(H)P(M) = (284/1976)(198/1976) = 0. P(E)P( S ) = (3/10)(3/10) = 0.025/50)) = 1 – (0.) = 1 – (1 – P(getting mal.3686 2-48.s. if seen in all firms. P(device works) = 1 – P(all components fail) = 1 – (1 – 0.))(1 – P(getting schist. 2-47.9476)(0.2202 The two events are not independent. 0.6667)(0.download full file at http://testbankcafe.000096. P(getting at least one disease) = 1 – P(getting none of the three) = 1 – P(not getting Mal.00048.com . (Use template: Probability of at least 1. There may be some sort of relationship between them as a general rule.09 P ( E  S ) = 2/10 = 0.999904)5 = 0. would be (0.xls) Probability of at least one success from many independent trials.9 Prob.8 0. (Use template: Probability of at least 1. assuming conditions stay the same as in 2003. download full file at http://testbankcafe.1382  0.))(1 – P(getting s.01468  0.)P(not getting s. 200 Permutation n 15 2-55.000096 0.25 0.5 0.459.7000 Since 1/4 of the items are in any particular quartile. nPr = n! / (n–r)! = 15! / (15-8)! = (15)(14)(13)(12)(11)(10)(9)(8) = 259.download full file at http://testbankcafe. P(all four in top quartile) = (1/4)4 = 1/256 = 0. of at least one success 0.000096 0.com Probability of at least one success from many independent trials. 0.50)(0.000096 0. 2-54.880 different orders.2 Prob.0039 P(at least one from bottom quartile) = 1 – P(all four from top three quartiles) = 1 – (3/4)4 = 175/256 = 0. and assuming independent random samples with replacement so that all four choices have the same probability of being in the top quartile. Success Probs 1 2 3 4 5 2-50.75)(0. (55)(30)(21)(13) = 450.684 2-52.70 Success Probs 1 2 3 2-51.000096 Prob. r 8 nPr 259459200 6P3 = 6! / (6–3)! = (6)(5)(4) = 120 ordered choices download full file at http://testbankcafe.30 = 0.com .80) = 1 – 0.0005 P(at least one drives home safely) = 1 – P(none drive home safely) = 1 – (0. of at least one success 0.000096 0.450 sets of representatives 2-53. 0. 9! = (9)(8)(7)(6)(5)(4)(3)(2)(1) = 362. 792 = 0. Let S. Only one possible combination of 3 elements chosen from the 14 parts consists of the 3 faulty ones.2) + (.25 P ( D | B ) = 0.86 2-62. P(D | B) = 0.45)(. So since any 3-element combination is equally likely to be picked.R be the events: successful takeover. Thus the probability of making such a guess is = 180/1.65 P(T | R) = 0.000000513 2-59.20 P( S ) = 0.B be the events: deal is concluded.37 2-63.com .6) = 0. 7! / [(7–2)!2!] = 21 pairs.792 = 0. W6} is the winning combination.9 P(S | E) = 0. competitor makes a bid..8 P(A) = P(A | S)P(S) + P(A | S )P( S ) = (.70 P(T) = P(T | R)P(R) + P(T | R)P(R) = (.0000924 2-60.6 P(D) = P(D | B)P(B) + P(D | B)P( B) = (. Let T.30)(.95 P(S) = 0.80) = 0.40 P(B ) = 0.95)(.download full file at http://testbankcafe.9)(.50 P(E) = 0. Let A. P(T | R) = 0.545 2-61. P(S | E) = 0.-.947.7) + (.7) + (. So (6) (30) = 180 possible combinations match exactly 5 of the winning numbers.50)(.25)(. then there are 6 choices of which wi is not in the guessed combination. and 30 possible wrong guesses in place of wi (since the one wrong guess can be any of the numbers from 1 to 36 that are not in W). Only one of the combinations wins.00275 2-58. resignation of a board member. .30) = 0.30) = 0. How many ways of guessing a set of 6 of the numbers from 1 to 36 will have 5 correct and 1 wrong? If W = {WI.S be the events: the drug is approved.57.78 download full file at http://testbankcafe.E be the events: property is sold.4) + (.947.30 P(R) = 0.45 P(B) = 0. Combination n r 7 2 nCr 21 2.5)(. W2.70 P(S) = P(S | E)P(E) + P(S | E)P(E) = (. the probability = 1 / (14! / 3!11!) = 1 / 364 = 0. P(A | S) = 0. the drug has side effects.65)(. economy improves. Let D.com 2-56.5 P(A | S ) = 0. so the probability of guessing it is 1 / (36! / 6!30!) = 1 / 1. 05 P(G | O) P(O) (.download full file at http://testbankcafe.O be the events: test indicates oil.9) + (.1) = 0.com . Let F.2857 2-68.15) = 0.92)(.4 P(I | O) = 0.15) P(H | I) = = P(I | H) P(H) + P(I |M) P(M) + P(I | L) P(L) (. low.G be the events: door should open. P(O) = 0.3 P(I | L) = 0.com 2-64.9944 2-66. Let I. green light appears.85)(.85)(.004) + (.75 P ( F | A = 0.1602 2-67.E be the events: the alarm sounds.75)(. Let S.85 P(I | O) = 0.M. Let I.7)+ (.23 P(F) = P(F | A)P(A) + P(F | A)P(A) = (.3)(.95)(.02 P(S | E) P(E) P(E | S) = (.02)(.77) = 0.L be the events: indicator rises.23) + (. P(I | H) = 0.98 P(G | O ) = 0.4) = P(I | O) P(O) + P(I | O )P( O ) (.1 P(H) = 0.15)+ (.4) + (.95 P(S | E) = 0.6 P(I | M) = 0.9) P(O | G) = = P(G | O) P(O) + P(G | O )P( O ) (.6) = 0.9 P(G | O) = 0. economic situation is high.996) = 0. there is an emergency situation.A be the events: ships sail full this summer. Let O. medium. oil really is present P(O) = 0.004 P(S | E) = 0.8809 2-65. P(E) = 0.05)(. P(F | A) = 0.6)(.98)(.95)(.10 P(I | O) P(O) P(O | I) = (.85 download full file at http://testbankcafe.1)(.7 P(L) = 0. dollar appreciates against European currencies.10)(.004) = P(S | E) P(E) + P(S | E)P(E) (.92 P(A) = 0.15 P(I | H) P(H) (.98)(.6)(.H.15 P(M) = 0. com download full file at http://testbankcafe.download full file at http://testbankcafe.com . 000 b) P(all leave) = (0.70. MF denote the events of reaching the two men.75)(. P(A) = 0. Let W be the event that a woman answers the door. P(  1 out of 5 misdirected) = 1 – P (none misdirected) = 1 – [P(a single call not misdirected)]5 = 1 – (199/200)5 = 0.15 + 0.4 P(J | C) = P(C  J) / P(C) = 0.6 P(I | S) = 0.15)(.4) = 0.3 – 0. P(C) = 0.S be the events: test indicates success.999 = approximately 1.60 download full file at http://testbankcafe.05 = 0.9 x 10-8 = approximately 0.12 P(B | A) = P(A  B) / P(A) = 0.20 = 0. the married couple (respectively). product really is successful.6 = 0. P(S) = 0.000 = 0. Let C.667 2-77. get job offer.000 c) P(at least one leaves) = 1 – (2.4/0.10 – 0.6) + (.02)(1/200) = 1/10.6 P(C  J) = 0.75)(. P(A  B) = 0.74.0001 2-73.15 P(I | S) P(S) (.download full file at http://testbankcafe. a) P(none leave) = (1 – 0. FF. Let I.com 2-69. the two women.8824 2.20 2-76.6 2.4 + 0.12/0. Let MM.75 P(I | S ) = 0. J be the events: pass CPA exam.6) P(S | I) = = P(I | S) P(S) + P(I | S )P( S ) (.9 x 10-8 = approximately 0. P(C  S) = 0.9 x 10-8 )= 0.20 P(A  B) = 0.5)25 = 2. Then: P(MF  W) = 1/6 P(MF) = 1/3 P(W | MF) = ½ P(MM) = 1/3 P(W | MM) = 0 P(MM  W) = 0 P(FF) = 1/3 P(W | FF) = 1 P(FF  W) = 1/3 P(W) = ½ P(MF | W) =  P(MF  W)/P(W) = [ (1/6) / (1/2) ] = 1/3 2-71.com .5)25 = 2. Assume independence in each manager’s decision.1 = 0.0 2-75. (0.0248 2-72. 6) = 0.35)(.download full file at http://testbankcafe.72 P(I) = 0.175  . I be the events: production increases. P ( EQ  SD ) = 0. Let S.15) = 0.3880 2-82.85 P ( S  R ) = P(R | S) P(S) = (.90)5 = 0.2 Prob. of at least one success 0.18 2-79. P(all 3 people consider EQ) = (. R be the events: see the ad in the Wall Street Journal. P(exposed to at least one mode of advertising) = 1 – P (not exposed to any of the 3) = 1 – (. Let EQ.com 2-78.0429 P(at least one person considers EQ) = 1 – P(none of the three consider EQ) = 1 – (. P(defective) = 0.50) = . The assumption of independence is justified by random sampling. sporty design rated among most important features. Let P.2)(1 – 0.85)(. P(P | I) = 0.7254 2-81.25 P(P  I) = P(P | I ) P(I) = (.35 P(SD) = 0.388 Success Probs 1 2 3 0.25.25) = 0. SD be the events: engineering quality.65)3 = 0.25 P(EQ) = 0.40951 2-84. P(S) = 0.51 51% of the people see and remember the advertisement 2-83.35)3 = 0.1 0.1 5 chips are chosen at random.72)(. remember it.80) = 0.50 (.85)(. therefore the two events are not independent 2-80. P(at least one color) = 1 – P(none of the colors) By independence: = 1 – (1– 0.90)(. interest rates decline more than half a point.6 P(R | S) = 0.59049 P(at least one is defective) = 1 – P(none are defective) = 0. P(none of the 5 are defective) = (0.com .524 download full file at http://testbankcafe.15 0.3)(1 – 0. of at least one success 0.6 P(N) = 0.15) + (0.1)(0.21 / 0.04)(.55)(0.75 P(A | L) = P(L  A) / P(L) = 0.65 The choice is probably not random. H be the events: customer defaults.04 P(H) = 0.2 0.2 P(U) = 0.30) = 0.132 2-86.21 = 0.6) + (0.6667 2-88.045 + 0.13)(.65) + (. Let D.com . political situation is favorable. N.0715 2-89.3)(0.1 P(F) = 0.15 Prob.13 P(D | H) = 0. P(S | F) = 0. Let S.5 P(L) = 0. P(D | H ) = 0.255 = 0. economy is high.3 0.2 P(S) = P(S | F) P(F) + P(S | N) P(N) + P(S | U)P(U) = (0.255 b) P(unskilled worker | assembly line worker) = 0.8235 download full file at http://testbankcafe.41 2-87.5/0.30)(0.5240 2-85.2) = 0. (2/3)5 = 0. a) P(assembly line worker) = P(skilled worker on assembly line) + P(unskilled worker on assembly line) = (0.com Success Probs 1 2 3 0. authorization is granted. neutral. Let L. U be the events: subsidiary will be successful. A be the events: legislation is passed.70)(0.3 P(S / U) = 0.75 = 0.download full file at http://testbankcafe. F. P(D) = P(D | H) P(H) + P(D | H)P(H) = (. P(L  A) = 0.2) + (0.35) = 0. unfavorable.55 P(S / N) = 0. Let S.35) + (.08 0. Let A.com . senior P(Se | A) P(A | Se) P(Se) = P(A | Se) P(Se) + P(A |F) P(F) + P(A | So) P(So) + P(A | J)P(J) (. C be the events: product is successful.08 0.65) + (.08 0.08 0. P(none are in error) = (0. Se be the events: student got an A.40)(. J.08 0. competitor produces similar product.8113 Success Probs 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 0.30)(.08 0. So.67 P(S | C) = 0.20) 2-93. sophomore.08 0.40)(.08 0.15) + (.08 Prob. P(S | C) = 0.08 0.com 2-90.2034 (.08 0.08 0.30) + (.download full file at http://testbankcafe.5825 download full file at http://testbankcafe. junior.08 0.08 0.95)10 = 0.35)(.08 0. P(at least one booking) = 1 – P(none) = 1 – (0.08 0.20)(.67)(.42)(. of at least one success 0.15) = = 0.35) = 0.08 0.08 0.42 P(C) = 0. student is a freshman. F.5987 2-92.35 P(S) = (.92)20 = 0.8113 2-91.08 0.08 0. removing one item does not appreciably alter the remaining number of data points on either side of the median). which is removed from play at the beginning. The probability of dying in a car crash in the U. The rate in the U. and their role in the outcome of the game should be more or less unnoticeable when averaged over a span of games. of at least one success 0.999855)20 = 0. Conceptually. 2-96. since half of the points in the population are on one side and half are on the other by the definition of median.08 0. Assume a large population so that the sampling can be considered as being done with replacement (i. The probability of not dying in a car crash in the U. the probabilities involved vary only slightly. Success Probs 1 2 3 4 5 6 7 8 9 10 11 12 13 0. So.6617 Probability of at least one success from many independent trials. 2-97.999855. over the next 20 years is: 1. download full file at http://testbankcafe.S.P(not dying) = 1 – (0.08 0.download full file at http://testbankcafe. Assume independence. however.com .S. for example.08 0.08 0. Thus the probability that the median will lie between the two points drawn is 1/2. she can't be sure of the probability that the next card is in fact smaller than a 4 (given the cards already drawn and known to everyone). while this space is always based on the full 52-card deck in the regular game.08 0.002896. if a player wants to make a draw and needs anything lower than a 4 in order not to go over 21. The probability of dying in a car crash in the U.08 0.e.08 Prob.08 0.6617 2-95.000096 (from problem 2-49). Practically speaking.S.5 times larger than the rate in France. since the hole card may or may not be one of the remaining cards smaller than 4.92)13 = 0.08 0.08 0.08 0.S. Then the first item drawn is on a particular side of the median.08 0. P(at least 1 orders drink) = 1 – P(none order drink) = 1 – (0. in the hole-card game there is a card missing from the space of cards available for play by the players-the hole card itself.94.08 0. In other words. The probability of dying in a car crash in France is 0. is 0. is 40676 / 280000000 = 0.000145. the probability space represented by the deck (in a randomized order) used for the main play isn't exactly specified in the hole-card game (from any player's point of view).. is 1. after which the second item has a 1/2 probability of coming from the other side.com 2. so the desired probability is 1 – (1/2). a simple majority would be 20 votes.0)(0. Here it is: the caller pre-selects one of H-T or T-H as the guessed sequence of outcomes of two consecutive flips.908146) . the caller attends the meeting iff his or her choice of the order of outcomes was correct. since the two tosses are independent. the sequence H-T is equally likely as T-H: they both have probability P(H) P(T) .9)][1 – (.. P(yes) = 0. = 1 – (. The median will lie somewhere between the smallest and largest values drawn exactly when the above situation does not occur.. open close total Kwik Save 893 107 1000 Somerfieldtotal 0 424 424 893 531 1424 a. 0. We need a method that is fair even if the caller knows which way the coin is biased (since otherwise a single random call would still have probability 1/2 of matching the actual outcome).00 c..14][1 – (.7 P(S) = P(S | G) P(G) + P(S | –G) P(–G) = (0. Since there are 38 members.0 download full file at http://testbankcafe.86)(.89794)(.58 P(S | –G) = 1. 107 / 531 = . 0 2-102.14)(.4)(0. The coin is then flipped twice: if it results in two different outcomes. = 1 – [1 – .202 b..0002 P(Northwest compete successfully) = 0.9)2] [1 – (.com 2-98.14)(.8866)(.874)(. S = Species Survives G = Project goes ahead Given Probabilities: P(–S | G) = 0. Using 800 iterations on a computer: = 0. repeated as needed until a pair of different outcomes is obtained.14)(.6 P(G) = 0.25 P(pass) = 0.com . 2-99.3) = 0. 2-101...766660928 2-100.7) + (1. then the coin is flipped twice again. the number of such ways is (1/2) n-1. But if the two flips have the same outcome. Making the same assumptions as in Problem 2-97. P (at least one paper accepted) = 1 – P (no papers accepted) = 1 – P(first paper rejected) P(second paper rejected) .9)3] .download full file at http://testbankcafe.089 = 1700 /19100 2-103.00014 2-104. first count the number of ways that all n elements drawn could lie on the same side of the median: since the choices are independent (this being a random sample) and each of the n – 1 choices after the first has probability 1/2 of being on the same side of the median as the first choice. 531 / 1424 + 1000 / 1424 – 107 / 1424 = 1. Note that even if the coin is much more likely to come up heads (and even if the caller knows this). 0208 I2 0. | I5) 1) Given the part is defective.38 0.86) = 1 – (0.36 Conditional Probabilities s1 defective P(I1 | .08 good P(I2 | .83) (0.xls) Bayesian Revision based on Empirical Conditional Probabilities Machine A Machine B Machine C s1 s2 s3 s4s5s6 Prior Probability 0. | I2) 0.92 P(I3 | .8.3654 s7 s8 s2 0.3838. It is better to apply to company 6.86)(0.2392 I3 I4 I5 Marginal 0.2529 P(. | I1) 0.9119 2) No.9458 Posterior Probabilities s1 P(.3610 s3 s4s5s6 0.82) (0.62) (0. the probability that it is from machine B is 0.) P(I5 | .3942) = 0.26 0. P(at least one offer) = 1 – P(no offers from 2.3817 s3 s4s5s6 0.com .2657 0. | I3) P(.3838 P(.9) = 1 – (0. 2) Given the part is good. Case 2: Job Applications 1) P(at least one offer) = 1 – P(no offers) = 1 – (0. but is cheaper.105.80) (0.0190 0.3506 0.65)(0. 3) $1430.0881 = 0.) 0.0542 0. (Use template: Bayesian Revision. the probability it is from machine A is 0.) P(I4 | .05 0.96 Joint Probabilities s1 I1 0.04 0. | I4) P(.86) (0.) 0.95 s3 1 1 Total 1 0.download full file at http://testbankcafe.92) = 1 – 0.6058 download full file at http://testbankcafe.com 2.3817.72) (0.) Total 1 s7 s8 s4s5s6 s7 s8 0 0 0 0 0 s2 0.5.82)(0.3456 s7 s8 s2 0.86) (0.0144 0.65) (0. since it has the same probability of success.82) (0. 9 or 10.com .7.8 or 9. download full file at http://testbankcafe.7644 probability of getting an offer from either company 2.3. 5) Through trial and error we arrive at a total cost of $1500 with a 0.download full file at http://testbankcafe.com 4) Through trial and error we arrive at a total cost of $2190 with a 0.8.6329 probability of getting an offer from either company 2.7.6. 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